Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct Equations {
  int a[10010], _[1000010], __[1000010], cnt, size, ins[30];
  Equations() : cnt(0) {}
  void addpath(int i, int j) { _[++cnt] = i, __[cnt] = j; }
  void Find_Linear_Space() {
    register int i, j, k;
    i = 1, j = 29;
    while (i <= size && j >= 0) {
      for (k = i; k <= size; ++k)
        if ((a[k] >> j) & 1) break;
      if ((a[k] >> j) & 1) {
        if (k != i) {
          a[k] ^= a[i], addpath(k, i);
          a[i] ^= a[k], addpath(i, k);
          a[k] ^= a[i], addpath(k, i);
        }
        ins[j] = i;
        for (k = 1; k <= size; ++k)
          if (k != i && (a[k] >> j) & 1) a[k] ^= a[i], addpath(k, i);
        ++i;
      }
      --j;
    }
  }
} x, y;
int main() {
  int n;
  scanf("%d", &n);
  x.size = y.size = n;
  register int i, j;
  for (i = 1; i <= n; ++i) scanf("%d", &x.a[i]);
  for (i = 1; i <= n; ++i) scanf("%d", &y.a[i]);
  x.Find_Linear_Space();
  y.Find_Linear_Space();
  for (j = 29; j >= 0; --j) {
    if (!x.ins[j]) continue;
    for (i = 1; i <= n; ++i)
      if (((x.a[i] ^ y.a[i]) >> j) & 1)
        x.a[i] ^= x.a[x.ins[j]], x.addpath(i, x.ins[j]);
  }
  bool find = 0;
  for (i = 1; i <= n; ++i)
    if (x.a[i] != y.a[i]) {
      find = 1;
      break;
    }
  if (find) {
    puts("-1");
    return 0;
  }
  printf("%d\n", x.cnt + y.cnt);
  for (i = 1; i <= x.cnt; ++i) printf("%d %d\n", x._[i], x.__[i]);
  for (i = y.cnt; i >= 1; --i) printf("%d %d\n", y._[i], y.__[i]);
  return 0;
}